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We will learn how to find the general values of inverse trigonometric functions in different types of problems.
1. Find the general values of sinβ1 (- β3/2)
Solution:
Let, sinβ1 (- β3/2) = ΞΈ
Therefore, sin ΞΈ = - β3/2
β sin ΞΈ = - sin (Ο/3)
β sin ΞΈ = (- Ο/3)
Therefore, the general value of sinβ1 (- β3/2) = ΞΈ = nΟ - (- 1)n Ο/3, where, n = 0 or any integer.
2.
Find the general values of cotβ1 (- 1)
Solution:
Let, cotβ1 (- 1) = ΞΈ
Therefore, cot ΞΈ = - 1
β cot ΞΈ = cot (- Ο/4)
Therefore, the general value of cotβ1 (- 1) = ΞΈ = nΟ - Ο/4, where, n = 0 or any integer.
3. Find the general values of cosβ1 (1/2)
Solution:
Let, cosβ1 1/2 = ΞΈ
Therefore, cos ΞΈ = 1/2
β cos ΞΈ = cos (Ο/3)
Therefore, the general value of cosβ1 (1/2) = ΞΈ = 2nΟ Β± Ο/3, where, n = 0 or any integer.
4. Find the general values of secβ1 (- 2)
Solution:
Let, secβ1 (- 2) = ΞΈ
Therefore, sec ΞΈ = - 2
β sec ΞΈ = - sec (Ο/3)
β sec ΞΈ = sec (Ο - Ο/3)
β sec ΞΈ = sec (2Ο/3)
Therefore, the general value of secβ1 (- 2) = ΞΈ = 2nΟ Β± 2Ο/3, where, n = 0 or any integer.
5. Find the general values of cscβ1 (β2)
Solution:
Let, cscβ1 (β2) = ΞΈ
Therefore, csc ΞΈ = β2 .
βcsc ΞΈ = csc (Ο/4)
Therefore, the general value of cscβ1 (β2 ) = ΞΈ = nΟ + (- 1)n Ο/4, where, n = 0 or any integer.
6. Find the general values of tanβ1 (β3)
Solution:
Let, tanβ1 (β3) = ΞΈ
Therefore, tan ΞΈ = β3
β tan ΞΈ = tan (Ο/3)
Therefore, the general value of tanβ1 (β3) = ΞΈ = nΟ + Ο/3 where, n = 0 or any integer.
β Inverse Trigonometric Functions
11 and 12 Grade Math
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